Find the solution(s) using the quadratic formula.
$x^2+6x-7=0$

• A. $x=2,x=-7$
• B. $x=1,x=-3$
• C. $x=1,x=-7$
• D. $x=2,x=-3$

Answer: (C)

Verified step by step guidance

1. Identify the coefficients in the quadratic equation \(x^2 + 6x - 7 = 0\). Here, \(a = 1\), \(b = 6\), and \(c = -7\).
2. Recall the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula gives the solutions to any quadratic equation \(ax^2 + bx + c = 0\).
3. Calculate the discriminant, which is the expression under the square root: \(\Delta = b^2 - 4ac\). Substitute the values to get \(\Delta = 6^2 - 4(1)(-7)\).
4. Evaluate the square root of the discriminant: \(\sqrt{\Delta} = \sqrt{36 + 28}\). This step determines whether the solutions are real and distinct, real and equal, or complex.
5. Substitute \(b\), \(a\), and \(\sqrt{\Delta}\) back into the quadratic formula to find the two solutions: \(x = \frac{-6 \pm \sqrt{64}}{2(1)}\). Simplify the numerator and denominator separately to express the two possible values for \(x\).